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If \( x=2+3 \cos \theta \) and \( y=1-3 \sin \theta \) represent a circle then the centre and radius is
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Verified Answer
The correct answer is:
\( (2,1), 3 \)
Given that,
$x=2+3 \cos \theta \rightarrow(1)$
$y=1-3 \sin \theta \rightarrow(2)$
From Eqs. $(1)$ and $(2)$, we have
$\frac{x-2}{3}=\cos \theta$ and $\frac{y-1}{-3}=\sin \theta$
We know that, $\sin ^{2} \theta+\cos ^{2} \theta=1$
So, $\left(\frac{x-2}{3}\right)^{2}+\left(\frac{y-1}{-3}\right)=1$
$\Rightarrow \frac{(x-2)^{2}}{9}+\frac{(y-1)^{2}}{9}=1$
$\Rightarrow(x-2)^{2}+(y-1)^{2}=9$
So, it is circle centre at point $(2,1)$ and radius is 3 .
$x=2+3 \cos \theta \rightarrow(1)$
$y=1-3 \sin \theta \rightarrow(2)$
From Eqs. $(1)$ and $(2)$, we have
$\frac{x-2}{3}=\cos \theta$ and $\frac{y-1}{-3}=\sin \theta$
We know that, $\sin ^{2} \theta+\cos ^{2} \theta=1$
So, $\left(\frac{x-2}{3}\right)^{2}+\left(\frac{y-1}{-3}\right)=1$
$\Rightarrow \frac{(x-2)^{2}}{9}+\frac{(y-1)^{2}}{9}=1$
$\Rightarrow(x-2)^{2}+(y-1)^{2}=9$
So, it is circle centre at point $(2,1)$ and radius is 3 .
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